OL ⊥ ON start overline, O, L, end overline, \perp, start overline, O, N, end overline \qquad m \angle LOM = 3x + 38^\circm∠LOM=3
x+38 ∘ m, angle, L, O, M, equals, 3, x, plus, 38, degrees \qquad m \angle MON = 9x + 28^\circm∠MON=9x+28 ∘ m, angle, M, O, N, equals, 9, x, plus, 28, degrees Find m\angle LOMm∠LOMm, angle, L, O, M:
If side OL is perpendicular to ON i.e OL ⊥ ON, then angle ∠LON = 90°. If the is another line OM projecting from O with ∠LOM= (3x+38)° and ∠MON= (9x+28)°, then ∠MON + ∠LOM = ∠LON
Substituting the given angles int the expressions above to calculate the value of x;
(9x+28)° + (3x+38)° = 90°
12x+66 = 90°
12x = 90-66
12x = 24
x = 2°
Since ∠LOM= (3x+38)°, to get the value of the angle, we will substitute x = 2° into the expression as shown;